Terminating balanced 4φ3の積公式

今回はterminating balanced${}_4\phi_3$の積を与える. まず, $q$-Saalschützの和公式 より$k\geq j$に対し,
\begin{align} \sum_{0\leq m}\frac{(q^{-j},q^{k-N},aq^{k+N};q)_m}{(q,aq^k,q^{1+k-j};q)_m}q^m&=\frac{(q^{-N},aq^N;q)_j}{(q^{-k},aq^k;q)_j} \end{align}
であるから,

\begin{align} &(q^{-N},aq^N;q)_j(q^{-N},aq^N;q)_k\\ &=(q^{-k},aq^k;q)_j\sum_{0\leq m}\frac{(q^{-j};q)_m(q^{-N},aq^{N};q)_{m+k}}{(q;q)_m(aq^k,q^{1+k-j};q)_m}q^m\\ &=(q;q)_j(q;q)_k(a;q)_{j+k}(-1)^jq^{\binom j2-jk}\sum_{0\leq m}\frac{(-1)^mq^{\binom m2-jm}(q^{-N},aq^{N};q)_{m+k}}{(q;q)_m(a;q)_{m+k}(q;q)_{m+k-j}(q;q)_{j-m}}q^m\\ &=(q;q)_j(q;q)_k(a;q)_{j+k}(-1)^jq^{\binom j2-jk}\sum_{0\leq m}\frac{(-1)^{k+m}q^{\binom {m-k}2-j(m-k)-k}(q^{-N},aq^{N};q)_{m}}{(a;q)_{m}(q;q)_{m-j}(q;q)_{m-k}(q;q)_{j+k-m}}q^{m}\\ &=(q;q)_j(q;q)_k(a;q)_{j+k}\sum_{0\leq m}\frac{(-1)^{j+k+m}q^{\binom {m}2+\binom{k}2+\binom j2-(j+k)m}(q^{-N},aq^{N};q)_{m}}{(a;q)_{m}(q;q)_{m-j}(q;q)_{m-k}(q;q)_{j+k-m}}q^{m} \end{align}
となるので, これを足し合わせて数列$B_j,C_k$に対して
\begin{align} &\sum_{j=0}^N\frac{(q^{-N},aq^N;q)_j}{(q,b;q)_j}B_j\sum_{k=0}^N\frac{(q^{-N},aq^N;q)_k}{(q,c;q)_k}C_k\\ &=\sum_{0\leq j,k}\frac{(a;q)_{j+k}}{(b;q)_j(c;q)_k}B_jC_k\sum_{0\leq m}\frac{(-1)^{j+k+m}q^{\binom {m}2+\binom{k}2+\binom j2-(j+k)m}(q^{-N},aq^{N};q)_{m}}{(a;q)_{m}(q;q)_{m-j}(q;q)_{m-k}(q;q)_{j+k-m}}q^{m}\\ &=\sum_{0\leq k}\frac{C_k}{(c;q)_k}\sum_{0\leq m}\frac{(-1)^{k+m}q^{\binom {m}2+\binom{k}2-km}(q^{-N},aq^{N};q)_{m}}{(a;q)_{m}(q;q)_{m-k}}q^{m}\sum_{0\leq j}\frac{(-1)^jq^{\binom j2-jm}(a;q)_{j+k}}{(b;q)_j(q;q)_{m-j}(q;q)_{j+k-m}}B_j\\ &=\sum_{0\leq k}\frac{C_k}{(c;q)_k}\sum_{0\leq m}\frac{q^{\binom {m}2+\binom{k}2}(q^{-N},aq^{N};q)_{m}}{(a;q)_{m}(q;q)_{m-k}}q^{m}\sum_{0\leq j}\frac{(-1)^jq^{\binom {m-k+j}2-(m+j)m}(a;q)_{j+m}}{(b;q)_{m-k+j}(q;q)_{k-j}(q;q)_j}B_{m-k+j}\\ &=\sum_{0\leq k}\frac{C_k}{(c,q;q)_k}\sum_{0\leq m}\frac{q^{-\binom {m}2+\binom{k}2+\binom {m-k}2}(q^{-N},aq^{N};q)_{m}}{(b,q;q)_{m-k}}\sum_{0\leq j}\frac{(q^{-k},aq^m;q)_j}{(bq^{m-k},q;q)_j}B_{m-k+j}\\ &=\sum_{0\leq m}(q^{-N},aq^{N};q)_{m}\sum_{0\leq k}\frac{q^{k^2-mk}}{(c,q;q)_k(b,q;q)_{m-k}}C_k\sum_{0\leq j}\frac{(q^{-k},aq^m;q)_j}{(bq^{m-k},q;q)_j}B_{m-k+j} \end{align}
つまり以下を得る.

補題1において
\begin{align} B_j&=\frac{(b_1,b_2;q)_j}{(b_3,ab_1b_2q/bb_3;q)_j}q^j,\qquad C_j=\frac{(c_1,c_2;q)_j}{(c_3,ac_1c_2q/cc_3;q)_j}q^j \end{align}
とすると
\begin{align} &\Q43{q^{-N},aq^N,b_1,b_2}{b,b_3,ab_1b_2q/bb_3}q\Q43{q^{-N},aq^N,c_1,c_2}{c,c_3,ac_1c_2q/cc_3}q\\ &=\sum_{0\leq m}(q^{-N},aq^{N};q)_{m}\sum_{0\leq k}\frac{q^{k^2-mk}}{(c,q;q)_k(b,q;q)_{m-k}}\frac{(c_1,c_2;q)_k}{(c_3,ac_1c_2q/cc_3;q)_k}q^k\\ &\qquad\cdot\sum_{0\leq j}\frac{(q^{-k},aq^m;q)_j}{(bq^{m-k},q;q)_j}\frac{(b_1,b_2;q)_{m-k+j}}{(b_3,ab_1b_2q/bb_3;q)_{m-k+j}}q^{m-k+j}\\ &=\sum_{0\leq m}(q^{-N},aq^{N};q)_{m}\sum_{0\leq k}\frac{q^{k^2-mk}(c_1,c_2;q)_{m-k}}{(b,q;q)_{k}(q,c,c_3,ac_1c_2q/cc_3;q)_{m-k}}q^{m}\\ &\qquad\cdot\sum_{0\leq j}\frac{(q^{k-m},aq^m;q)_j}{(bq^{k},q;q)_j}\frac{(b_1,b_2;q)_{j+k}}{(b_3,ab_1b_2q/bb_3;q)_{j+k}}q^{j}\qquad k\mapsto m-k\\ &=\sum_{0\leq m}(q^{-N},aq^{N};q)_{m}\sum_{0\leq k}\frac{q^{k^2-mk}(c_1,c_2;q)_{m-k}}{(b,q;q)_{k}(q,c,c_3,ac_1c_2q/cc_3;q)_{m-k}}q^{m}\\ &\qquad\cdot\frac{(b_1,b_2;q)_k}{(b_3,ab_1b_2q/bb_3;q)_k}\Q43{q^{k-m},aq^m,b_1q^k,b_2q^k}{bq^k,b_3q^k,ab_1b_2q^{k+1}/bb_3}q \end{align}
ここで, Watsonの変換公式 より
\begin{align} &\Q43{q^{k-m},aq^m,b_1q^k,b_2q^k}{bq^k,b_3q^k,ab_1b_2q^{k+1}/bb_3}q\\ &=\frac{(bb_3q^{k-m}/ab_1,bb_3q^{k-m}/ab_2;q)_{m-k}}{(bb_3q^{2k-m}/a,bb_3q^{-m}/ab_1b_2;q)_{m-k}}\\ &\qquad\cdot\Q87{bb_3q^{2k-m-1}/a,q\sqrt{bb_3q^{2k-m-1}/a},-q\sqrt{bb_3q^{2k-m-1}/a},b_3q^{k-m}/a,bq^{k-m}/a,b_1q^k,b_2q^k,q^{k-m}}{\sqrt{bb_3q^{2k-m-1}/a},-\sqrt{bb_3q^{2k-m-1}/a},bq^k,b_3q^k,bb_3q^{k-m}/ab_1,bb_3q^{k-m}/ab_2,bb_3q^k/a}{\frac{bb_3q^{m-k}}{b_1b_2}} \end{align}
となるので, これを代入して,
\begin{align} &\sum_{0\leq m}(q^{-N},aq^{N};q)_{m}\sum_{0\leq k}\frac{q^{k^2-mk}(c_1,c_2;q)_{m-k}}{(b,q;q)_{k}(q,c,c_3,ac_1c_2q/cc_3;q)_{m-k}}q^{m}\\ &\qquad\cdot\frac{(b_1,b_2;q)_k}{(b_3,ab_1b_2q/bb_3;q)_k}\Q43{q^{k-m},aq^m,b_1q^k,b_2q^k}{bq^k,b_3q^k,ab_1b_2q^{k+1}/bb_3}q\\ &=\sum_{0\leq m}(q^{-N},aq^{N};q)_{m}\sum_{0\leq k}\frac{q^{k^2-mk}(c_1,c_2;q)_{m-k}}{(b,q;q)_{k}(q,c,c_3,ac_1c_2q/cc_3;q)_{m-k}}q^{m}\\ &\qquad\cdot\frac{(b_1,b_2;q)_k}{(b_3,ab_1b_2q/bb_3;q)_k}\frac{(bb_3q^{k-m}/ab_1,bb_3q^{k-m}/ab_2;q)_{m-k}}{(bb_3q^{2k-m}/a,bb_3q^{-m}/ab_1b_2;q)_{m-k}}\\ &\qquad\cdot\sum_{0\leq j}\frac{1-bb_3q^{2j+2k-m-1}/a}{1-bb_3q^{2k-m-1}/a}\frac{(bb_3q^{2k-m-1}/a,b_3q^{k-m}/a,bq^{k-m}/a,b_1q^k,b_2q^k,q^{k-m};q)_j}{(q,bq^k,b_3q^k,bb_3q^{k-m}/ab_1,bb_3q^{k-m}/ab_2,bb_3q^k/a;q)_j}\left(\frac{bb_3q^{m-k}}{b_1b_2}\right)^j\\ &=\sum_{0\leq m}(q^{-N},aq^{N};q)_{m}\sum_{0\leq k\leq m}\frac{q^{k^2-mk}(c_1,c_2;q)_{m-k}}{(b,q;q)_{k}(q,c,c_3,ac_1c_2q/cc_3;q)_{m-k}}q^{m}\\ &\qquad\cdot\frac{(b_1,b_2;q)_k}{(b_3,ab_1b_2q/bb_3;q)_k}\frac{(bb_3q^{k-m}/ab_1,bb_3q^{k-m}/ab_2;q)_{m-k}}{(bb_3q^{2k-m}/a,bb_3q^{-m}/ab_1b_2;q)_{m-k}}\frac{(b,b_3,bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,bb_3/a;q)_k}{(bb_3q^{-m}/a;q)_{2k}(b_3q^{-m}/a,bq^{-m}/a,b_1,b_2,q^{-m};q)_k}\\ &\qquad\cdot\sum_{0\leq j}\frac{1-bb_3q^{2j+2k-m-1}/a}{1-bb_3q^{-m-1}/a}\frac{(bb_3q^{-m-1}/a;q)_{j+2k}(b_3q^{-m}/a,bq^{-m}/a,b_1,b_2,q^{-m};q)_{j+k}}{(q;q)_j(b,b_3,bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,bb_3/a;q)_{j+k}}\left(\frac{bb_3q^{m-k}}{b_1b_2}\right)^j\\ &=\sum_{0\leq m}\frac{(q^{-N},aq^{N};q)_{m}}{(q;q)_m}q^{m}\sum_{0\leq k\leq m}\frac{(-1)^kq^{\binom{k+1}2}(c_1,c_2;q)_{m-k}}{(q;q)_{k}(c,c_3,ac_1c_2q/cc_3;q)_{m-k}}\\ &\qquad\cdot\frac{(bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2;q)_{m}}{(bb_3q^{-m}/a;q)_{m+k}(bb_3q^{-m}/ab_1b_2;q)_{m-k}}\frac{(bb_3/a;q)_k}{(ab_1b_2q/bb_3,b_3q^{-m}/a,bq^{-m}/a;q)_k}\\ &\qquad\cdot\sum_{0\leq j}\frac{1-bb_3q^{2j+2k-m-1}/a}{1-bb_3q^{-m-1}/a}\frac{(bb_3q^{-m-1}/a;q)_{j+2k}(b_3q^{-m}/a,bq^{-m}/a,b_1,b_2,q^{-m};q)_{j+k}}{(q;q)_j(b,b_3,bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,bb_3/a;q)_{j+k}}\left(\frac{bb_3q^{m-k}}{b_1b_2}\right)^j\\ &=\sum_{0\leq m}\frac{(q^{-N},aq^{N},bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,c_1,c_2;q)_{m}}{(q,bb_3q^{-m}/a,bb_3q^{-m}/ab_1b_2,c,c_3,ac_1c_2q/cc_3;q)_m}q^{m}\\ &\qquad\cdot\sum_{0\leq k}\frac{(-1)^kq^{\binom{k+1}2}(c_1q^m,c_2q^m;q)_{-k}}{(q,ab_1b_2q/bb_3,b_3q^{-m}/a,bq^{-m}/a;q)_{k}(cq^m,c_3q^m,ac_1c_2q^{m+1}/cc_3,bb_3/ab_1b_2;q)_{-k}}\\ &\qquad\cdot\sum_{0\leq j}\frac{1-bb_3q^{2j+2k-m-1}/a}{1-bb_3q^{-m-1}/a}\frac{(bb_3q^{-m-1}/a;q)_{j+2k}(b_3q^{-m}/a,bq^{-m}/a,b_1,b_2,q^{-m};q)_{j+k}}{(q;q)_j(b,b_3,bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,bb_3/a;q)_{j+k}}\left(\frac{bb_3q^{m-k}}{b_1b_2}\right)^j\\ &=\sum_{0\leq m}\frac{(q^{-N},aq^{N},ab_1q/bb_3,ab_2q/bb_3,c_1,c_2;q)_{m}}{(q,aq/bb_3,ab_1b_2q/bb_3,c,c_3,ac_1c_2q/cc_3;q)_m}q^{m}\\ &\qquad\cdot\sum_{0\leq k}\frac{(-1)^kq^{\binom k2}(q^{1-m}/c,q^{1-m}/c_3,cc_3q^{-m}/ac_1c_2;q)_k}{(q,b_3q^{-m}/a,bq^{-m}/a,q^{1-m}/c_1,q^{1-m}/c_2;q)_{k}}q^k\\ &\qquad\cdot\sum_{k\leq j}\frac{1-bb_3q^{2j-m-1}/a}{1-bb_3q^{-m-1}/a}\frac{(bb_3q^{-m-1}/a;q)_{j+k}(b_3q^{-m}/a,bq^{-m}/a,b_1,b_2,q^{-m};q)_{j}}{(q;q)_{j-k}(b,b_3,bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,bb_3/a;q)_{j}}\left(\frac{bb_3q^{m-k}}{b_1b_2}\right)^j\\ &=\sum_{0\leq m}\frac{(q^{-N},aq^{N},ab_1q/bb_3,ab_2q/bb_3,c_1,c_2;q)_{m}}{(q,aq/bb_3,ab_1b_2q/bb_3,c,c_3,ac_1c_2q/cc_3;q)_m}q^{m}\\ &\qquad\cdot\sum_{0\leq j}\frac{1-bb_3q^{2j-m-1}/a}{1-bb_3q^{-m-1}/a}\frac{(bb_3q^{-m-1}/a,b_3q^{-m}/a,bq^{-m}/a,b_1,b_2,q^{-m};q)_{j}}{(q,b,b_3,bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,bb_3/a;q)_{j}}\left(\frac{bb_3q^{m}}{b_1b_2}\right)^j\\ &\qquad\cdot\sum_{0\leq k}\frac{(q^{-j},q^{1-m}/c,q^{1-m}/c_3,cc_3q^{-m}/ac_1c_2,bb_3q^{j-m-1}/a;q)_k}{(q,b_3q^{-m}/a,bq^{-m}/a,q^{1-m}/c_1,q^{1-m}/c_2;q)_{k}}q^k \end{align}
よって,
\begin{align} &\Q43{q^{-N},aq^N,b_1,b_2}{b,b_3,ab_1b_2q/bb_3}q\Q43{q^{-N},aq^N,c_1,c_2}{c,c_3,ac_1c_2q/cc_3}q\\ &=\sum_{0\leq m}\frac{(q^{-N},aq^{N},ab_1q/bb_3,ab_2q/bb_3,c_1,c_2;q)_{m}}{(q,aq/bb_3,ab_1b_2q/bb_3,c,c_3,ac_1c_2q/cc_3;q)_m}q^{m}\\ &\qquad\cdot\sum_{0\leq j}\frac{1-bb_3q^{2j-m-1}/a}{1-bb_3q^{-m-1}/a}\frac{(bb_3q^{-m-1}/a,b_3q^{-m}/a,bq^{-m}/a,b_1,b_2,q^{-m};q)_{j}}{(q,b,b_3,bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,bb_3/a;q)_{j}}\left(\frac{bb_3q^{m}}{b_1b_2}\right)^j\\ &\qquad\cdot\Q54{q^{-j},q^{1-m}/c,q^{1-m}/c_3,cc_3q^{-m}/ac_1c_2,bb_3q^{j-m-1}/a}{b_3q^{-m}/a,bq^{-m}/a,q^{1-m}/c_1,q^{1-m}/c_2}q \end{align}
を得る. 特に$c=aq/b,c_3=aq/b_3$のとき, $q$-Saalschützの和公式 より
\begin{align} &\Q43{q^{-N},aq^N,b_1,b_2}{b,b_3,ab_1b_2q/bb_3}q\Q43{q^{-N},aq^N,c_1,c_2}{aq/b,aq/b_3,bb_3c_1c_2/aq}q\\ &=\sum_{0\leq m}\frac{(q^{-N},aq^{N},ab_1q/bb_3,ab_2q/bb_3,c_1,c_2;q)_{m}}{(q,aq/bb_3,ab_1b_2q/bb_3,aq/b,aq/b_3,bb_3c_1c_2/aq;q)_m}q^{m}\\ &\qquad\cdot\sum_{0\leq j}\frac{1-bb_3q^{2j-m-1}/a}{1-bb_3q^{-m-1}/a}\frac{(bb_3q^{-m-1}/a,b_3q^{-m}/a,bq^{-m}/a,b_1,b_2,q^{-m};q)_{j}}{(q,b,b_3,bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,bb_3/a;q)_{j}}\left(\frac{bb_3q^{m}}{b_1b_2}\right)^j\\ &\qquad\cdot\Q32{q^{-j},aq^{2-m}/bb_3c_1c_2,bb_3q^{j-m-1}/a}{q^{1-m}/c_1,q^{1-m}/c_2}q\\ &=\sum_{0\leq m}\frac{(q^{-N},aq^{N},ab_1q/bb_3,ab_2q/bb_3,c_1,c_2;q)_{m}}{(q,aq/bb_3,ab_1b_2q/bb_3,aq/b,aq/b_3,bb_3c_1c_2/aq;q)_m}q^{m}\\ &\qquad\cdot\sum_{0\leq j}\frac{1-bb_3q^{2j-m-1}/a}{1-bb_3q^{-m-1}/a}\frac{(bb_3q^{-m-1}/a,b_3q^{-m}/a,bq^{-m}/a,b_1,b_2,q^{-m};q)_{j}}{(q,b,b_3,bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,bb_3/a;q)_{j}}\left(\frac{bb_3q^{m}}{b_1b_2}\right)^j\\ &\qquad\cdot\frac{(bb_3c_1/aq,bb_3c_2/aq;q)_j}{(q^{1-m}/c_1,q^{1-m}/c_2;q)_j}\left(\frac{aq^{2-m}}{bb_3c_1c_2}\right)^j\\ &=\sum_{0\leq m}\frac{(q^{-N},aq^{N},ab_1q/bb_3,ab_2q/bb_3,c_1,c_2;q)_{m}}{(q,aq/bb_3,ab_1b_2q/bb_3,aq/b,aq/b_3,bb_3c_1c_2/aq;q)_m}q^{m}\\ &\qquad\cdot\Q{10}{9}{bb_3q^{-m-1}/a,q\sqrt{bb_3q^{-m-1}/a},-q\sqrt{bb_3q^{-m-1}/a},b_1,b_2,b_3q^{-m}/a,bq^{-m}/a,bb_3c_1/aq,bb_3c_2/aq,q^{-m}}{\sqrt{bb_3q^{-m-1}/a},-\sqrt{bb_3q^{-m-1}/a},bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,b_3,b,q^{1-m}/c_1,q^{1-m}/c_2,bb_3/a}{\frac{aq^2}{b_1b_2c_1c_2}} \end{align}
つまり, 以下を得る.

これは Searsの変換公式 より
\begin{align} \Q43{q^{-N},aq^N,c_1,c_2}{aq/b,aq/b_3,bb_3c_1c_2/aq}q&=\frac{(q^{1-N}/b,q^{1-N}/b_3;q)_N}{(aq/b,aq/b_3;q)_N}(aq^N)^N\Q43{q^{-N},aq^N,bb_3c_1/aq,bb_3c_2/aq}{b,b_3,bb_3c_1c_2/aq}{q}\\ &=\frac{(b,b_3;q)_N}{(aq/b,aq/b_3;q)_N}\left(\frac{aq}{bb_3}\right)^N\Q43{q^{-N},aq^N,bb_3c_1/aq,bb_3c_2/aq}{b,b_3,bb_3c_1c_2/aq}{q} \end{align}
であるから,
\begin{align} &\frac{(b,b_3;q)_N}{(aq/b,aq/b_3;q)_N}\left(\frac{aq}{bb_3}\right)^N\Q43{q^{-N},aq^N,b_1,b_2}{b,b_3,ab_1b_2q/bb_3}q\Q43{q^{-N},aq^N,bb_3c_1/aq,bb_3c_2/aq}{b,b_3,bb_3c_1c_2/aq}{q}\\ &=\sum_{0\leq m}\frac{(q^{-N},aq^{N},ab_1q/bb_3,ab_2q/bb_3,c_1,c_2;q)_{m}}{(q,aq/bb_3,ab_1b_2q/bb_3,aq/b,aq/b_3,bb_3c_1c_2/aq;q)_m}q^{m}\\ &\qquad\cdot\Q{10}{9}{bb_3q^{-m-1}/a,q\sqrt{bb_3q^{-m-1}/a},-q\sqrt{bb_3q^{-m-1}/a},b_1,b_2,b_3q^{-m}/a,bq^{-m}/a,bb_3c_1/aq,bb_3c_2/aq,q^{-m}}{\sqrt{bb_3q^{-m-1}/a},-\sqrt{bb_3q^{-m-1}/a},bb_3q^{-m}/ab_1,bb_3q^{-m}/ab_2,b_3,b,q^{1-m}/c_1,q^{1-m}/c_2,bb_3/a}{\frac{aq^2}{b_1b_2c_1c_2}} \end{align}
と書き換えることもできる. ここで, $b_2\mapsto bb_2, c_1\mapsto c_1/b$としてから$b\to 0$とすると,
\begin{align} &\frac{(b_3;q)_N}{(aq/b_3;q)_N}(-b_3)^{-N}q^{-\binom N2}\Q32{q^{-N},aq^N,b_1}{b_3,ab_1b_2q/b_3}q\Q32{q^{-N},aq^N,b_3c_1/aq}{b_3,b_3c_1c_2/aq}{q}\\ &=\sum_{0\leq m}\frac{(q^{-N},aq^{N},ab_2q/b_3,c_2;q)_{m}}{(q,ab_1b_2q/b_3,aq/b_3,b_3c_1c_2/aq;q)_m}\left(\frac{b_1c_1}a\right)^{m}\\ &\qquad\cdot\Q43{b_1,b_3q^{-m}/a,b_3c_1/aq,q^{-m}}{b_3q^{-m}/ab_2,b_3,q^{1-m}/c_2}{\frac{aq^2}{b_1b_2c_1c_2}}\\ &=\sum_{0\leq m}\frac{(q^{-N},aq^{N};q)_{m}}{(ab_1b_2q/b_3,b_3c_1c_2/aq;q)_m}\left(\frac{b_1c_1}a\right)^{m}\\ &\qquad\cdot\sum_{k=0}^m\frac{(b_1,b_3c_1/aq;q)_k(ab_2q/b_3,c_2;q)_{m-k}}{(q,b_3;q)_k(q,aq/b_3;q)_{m-k}}\left(\frac{aq}{b_1c_1}\right)^k \end{align}
となる. 変数を置き換えると以下を得る.

これは Gasperによって示された公式
\begin{align} &\frac{(-1)^N(c)_N}{(1+a-c)_N}\F32{-N,a+N,b}{c,d}1\F32{-N,a+N,e}{c,f}1\\ &=\sum_{0\leq k}\frac{(-N,a+N)_k}{(d,f)_k}\sum_{j=0}^k\frac{(b,e)_j}{j!(c)_j}\frac{(d-b,f-e)_{k-j}}{(k-j)!(1+a-c)_{k-j}} \end{align}
の$q$類似である. さらに$d=f=0, b\mapsto x,e\mapsto y$とすると
\begin{align} &\frac{(c;q)_N}{(aq/c;q)_N}(-c)^{-N}q^{-\binom N2}\Q32{q^{-N},aq^N,x}{c,0}q\Q32{q^{-N},aq^N,y}{c,0}{q}\\ &=\sum_{0\leq m}(q^{-N},aq^{N};q)_{m}\left(\frac{xyq}c\right)^{m}\sum_{k=0}^m\frac{(x,y;q)_k}{(q,c;q)_k(q,aq/c;q)_{m-k}}\left(\frac{c}{xy}\right)^k\\ &=\sum_{0\leq m,k}\frac{(q^{-N},aq^{N};q)_{m+k}}{(q,c;q)_k(q,aq/c;q)_{m}}\left(\frac{xyq}c\right)^{m}(x,y;q)_kq^k \end{align}
つまり, 以下を得る.

これは Watsonによる積公式
\begin{align} &\frac{(-1)^N(c)_N}{(1+a-c)_N}\F21{-N,a+N}{c}{1-x}\F21{-N,a+N}{c}{1-y}\\ &=\F{}4{-N,a+N}{c,1+a-c}{(1-x)(1-y),xy} \end{align}
の$q$類似である.